Let $y^4-2x=5$. What is the value of $\dfrac{d^2y}{dx^2}$ at the point $(-2,1)$ ? Give an exact number.
Answer: Notice that the equation defines $y$ implicitly—we don't have an explicit expression for $y$ in terms of $x$. So we will have to use implicit differentiation. If we differentiate the equation once, we will be able to get an expression for $\dfrac{dy}{dx}$. Then we can differentiate the equation again to get an expression for $\dfrac{d^2y}{dx^2}$. Let's start by finding $\dfrac{dy}{dx}$. $\dfrac{dy}{dx}=\dfrac{1}{2y^3}$ Now we can differentiate $\dfrac{dy}{dx}$ to find $\dfrac{d^2y}{dx^2}$. $\dfrac{d^2y}{dx^2}=-\dfrac{3}{4y^7}$ Finally, let's plug ${x=-2}$ and ${y=1}$ into the expression we got: $\begin{aligned} \left.-\dfrac{3}{4 y^7}\right\rvert_{({-2},{1})}&=-\dfrac{3}{4({1})^7} \\\\ &=-\dfrac{3}{4} \end{aligned}$ (Notice that the expression for $\dfrac{d^2y}{dx^2}$ only depends on $ y$ so there was no need to plug ${x=-2}$.) In conclusion, the value of $\dfrac{d^2y}{dx^2}$ at the point $(-2,1)$ is $-\dfrac{3}{4}$.